(a) Explanation Calculation: The given equation is d 2 y d x 2 − 4 d y d x − 5 y = e x + 4 . Substitution of y = e r x into the differential equation yields the auxiliary equation r 2 − 4 r − 5 = 0 . The equation r 2 − 4 r − 5 = 0 is factored as follows. r 2 − 4 r − 5 = 0 r 2 − 5 r + r − 5 = 0 r ( r − 5 ) + 1 ( r − 5 ) = 0 ( r + 1 ) ( r − 5 ) = 0 Here, the roots are r 1 = 5 and r 2 = − 1 . It is cleared that r 1 and r 2 are two real and unequal roots of the auxiliary equation r 2 − 4 r − 5 = 0 , then y c = c 1 e 5 x + c 2 e − x is the general solution to d 2 y d x 2 − 4 d y d x − 5 y = 0 . Since y 1 ( x ) = e 5 x and y 2 ( x ) = e − x , the conditions to be satisfied as, v 1 ′ e 5 x + v 2 ′ e − x = 0 v 1 ′ 5 e 5 x − v 2 ′ e − x = e x + 4 Solving the above system for v 1 ′ and v 2 ′ gives, v 1 ′ = | 0 e − x e x + 4 − e − x | | e 5 x e − x 5 e 5 x − e − x | = − 1 − 4 e − x − 6 e 4 x = 1 6 e − 4 x + 2 3 (b) To determine To solve: The equation d 2 y d x 2 − 4 d y d x − 5 y = e x + 4 by the method of undetermined coefficients.

Thomas' Calculus: Early Transcendentals plus MyLab Math with Pearson eText -- Title-Specific Access Card Package (14th Edition)

14th Edition

ISBN: 9780134768496

Author: Joel R. Hass, Christopher E. Heil, Maurice D. Weir

Publisher: PEARSON

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Chapter 17.2, Problem 35E

(a)

To determine

To solve: The equation d2ydx2−4dydx−5y=ex+4 by variation of parameters.

(b)

To determine

To solve: The equation d2ydx2−4dydx−5y=ex+4 by the method of undetermined coefficients.

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Chapter 17.2, Problem 34E

Chapter 17.2, Problem 36E

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Thomas' Calculus: Early Transcendentals plus MyLab Math with Pearson eText -- Title-Specific Access Card Package (14th Edition)

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The equation d 2 y d x 2 − 4 d y d x − 5 y = e x + 4 by variation of parameters.